---
title: "ILD for bending moment at section D of arch"
subject: "Theory of Structures I"
chapter: 6
key: "semester_4/theory_of_structures_i:arch-ild-bm-d-given-arch"
source: https://www.bctengineer.com/q/semester_4/theory_of_structures_i/arch-ild-bm-d-given-arch
site: BCT Engineering
---

# ILD for bending moment at section D of arch

Theory of Structures I · Chapter 6: Statically Determinate Arches

- Asked in: 2079 Chaitra, 6 marks
- Video: [Influence Line Diagram Moving Load Part II |  Bending Moment in Section | Theory of Structure I](https://www.youtube.com/watch?v=ZDVVY0TGLC0) — Parash Joshi - Civil Construction and Tutor
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_4/theory_of_structures_i/arch-ild-bm-d-given-arch
- Whole chapter: https://www.bctengineer.com/md/semester_4/theory_of_structures_i/chapter-6

## Question

Draw ILD for bending moment at section D for the given arch. [Figure: three hinged circular arch ACB, span 60 m, with crown hinge C, section D on the left half; see the loading in the companion question.]

## Short answer

The ILD for bending moment at section D (15 m from A) is a straight line from zero at A to +5.586 m at D, crosses zero at the neutral point 23.9 m from A, reaches -3.827 m at the crown, and returns to zero at B. Loads left of the neutral point produce positive moment; loads right produce negative moment.

## Answer

**Data:** three-hinged circular arch, $L = 60$ m, $f = 5$ m (radius $R = \dfrac{900+25}{10} = 92.5$ m), crown hinge $C$ at 30 m. Section $D$ in the left half at $x = 15$ m from $A$ (as in the loaded arch of the companion problem): $y_D = 3.776$ m.

#### Principle
For a unit load at $x'$:

- $R_A = \dfrac{L-x'}{L}$, thrust $H = \dfrac{M_C}{f}$ with $M_C = \dfrac{x'}{2}$ (load on the left half) or $\dfrac{L-x'}{2}$ (right half).
- $M_D = R_A\,x_D - H\,y_D$ for a load right of $D$, and $M_D = R_B\,(L-x_D) - H\,y_D$ for a load left of $D$.

$$
H = \frac{x'}{2f} = \frac{x'}{10} \ (x'\le 30), \qquad \frac{60-x'}{10}\ (x'\ge 30)
$$

#### Ordinates
For a load at $D$ itself ($x'=15$): $M = \dfrac{15\times 45}{60} - \dfrac{15}{10}\times 3.776 = 5.586$ m. At the crown ($x'=30$): $M = 0.5\times 15 - 3\times 3.776 = -3.827$ m.

| x (m) | Point | ILD of H | ILD of M at section |
|---|---|---|---|
| 0 | A | 0 | 0 |
| 15 | section D | 1.5 | 5.586 |
| 23.9 | neutral point | 2.39 | 0 |
| 30 | crown C | 3 | -3.827 |
| 60 | B | 0 | 0 |

#### Neutral point
$$
s_0 = \frac{2f}{\dfrac{2f}{L}+\dfrac{y_D}{x_D}} = \frac{10}{\dfrac{10}{60}+\dfrac{3.776}{15}} = 23.9\ \text{m from } A
$$

#### Shape of the ILD
```text
  ILD of M_D (m), straight between the points:
   A = 0  ->  D = +5.586  ->  N = 0  ->  C = -3.827  ->  B = 0
```

The ILD of $M_D$ is a straight line from 0 at $A$ to $+5.586$ m at $D$, then a straight line crossing zero at the neutral point (23.9 m), reaching $-3.827$ m at the crown and returning in a straight line to 0 at $B$. A load between $A$ and the neutral point produces positive (sagging) moment at $D$; a load between the neutral point and $B$ produces negative moment.

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. For L = 60 m and f = 5 m, what is H for a unit load at x' <= 30?
   A. x'/10
   B. x'/5
   C. x'/60
   D. x'/20
   Answer: A. H = M_C/f = (x'/2)/5 = x'/10. (Key point: Thrust ILD)

2. What is the ILD ordinate of M at D when the unit load is at D (x_D = 15 m, y_D = 3.776)?
   A. 5.586 m
   B. 3.827 m
   C. 11.25 m
   D. 0
   Answer: A. M = 15 x 45/60 - 1.5 x 3.776 = 11.25 - 5.664 = 5.586. (Key point: Ordinate at D)

3. Where is the neutral point of the M_D ILD?
   A. 23.9 m from A
   B. 30 m from A
   C. 15 m from A
   D. 45 m from A
   Answer: A. s0 = 10/(10/60 + 3.776/15) = 23.9 m. (Key point: Neutral point)

4. A load between the neutral point and B causes what moment at D?
   A. Positive
   B. Negative (hogging), e.g. -3.827 at the crown
   C. Zero
   D. Infinite
   Answer: B. The ILD crosses zero at the neutral point and is negative beyond it. (Key point: Sign beyond neutral point)

## Related in Theory of Structures I

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## Related in other subjects

- [ILD: bending moment at B, two-span beam](https://www.bctengineer.com/question.md?key=semester_5%2Ftheory_of_structures_ii%3Aild-bm-b-two-span-8) — Theory of Structures II, chapter 4
- [Two-hinged arch: BM at section x-x](https://www.bctengineer.com/question.md?key=semester_5%2Ftheory_of_structures_ii%3Aarch-bm-section-xx-60-85) — Theory of Structures II, chapter 2
- [ILD: bending moment at D, two-span beam](https://www.bctengineer.com/question.md?key=semester_5%2Ftheory_of_structures_ii%3Aild-bm-d-ab-bc-2) — Theory of Structures II, chapter 4
- [Two-hinged arch: BM, thrust, radial shear, support yield](https://www.bctengineer.com/question.md?key=semester_5%2Ftheory_of_structures_ii%3Aarch-bm-thrust-shear-yield-support) — Theory of Structures II, chapter 2